For the program below:
public class ZeroFillRightShift
{
public static void main(String args[])
{
int x = -1;
int y = x>>>1;
System.out.println("x = " + x);
System.out.println("y = " + y);
}
I get the output as follows:
x = -1
y = 2147483647
The result that I got for -1>>>1
is 2147483647. If it’s the sign bit that has to be shifted, as I learned, the result should be 1073741824. Why is it 2147483647 then?
The following image illustrates my problem more clearly:
The unsigned right shift operator ">>>" shifts a zero into the leftmost position, while the leftmost position after ">>" depends on sign extension.
So, -1 is shifted right one bit with zero-extension, which means it will insert a 0 into the leftmost position. Remember, we're dealing with two's complement here:
-1 is: 11111111111111111111111111111111
or 0xFFFFFFFF
in Hex
-1 >>> 1 is 01111111111111111111111111111111
or 0x7FFFFFFF
in Hex,
which is 231 - 1 == 2147483647
Here's a JLS reference for shift operators.
You seemed to be confused about two's complement. 31 bits are used for the value and the bit farthest to the left is used for the sign. Since you're only shifting by 1 bit, the signed bit becomes a 0, which means positive, and the result is the greatest positive number than an int
can represent.
Perhaps another example will help. Let's consider the following:
System.out.println(-2 >> 1); //prints -1
-2 = 11111111111111111111111111111110
If we use a signed right shift, we get: 11111111111111111111111111111111
, which is -1. However, if we do:
System.out.println(-2 >>> 1); //prints 2147483647
since -2 = 11111111111111111111111111111110
and do an unsigned right shift, which means we shift by 1 bit with zero-extension, giving: 01111111111111111111111111111111
The 32-bit decimal (two's complement) value -1 is 0xFFFFFFFF in hexadecimal. If you do an unsigned right shift ( >>>
) by one bit on that you get 0x7FFFFFFF, which is 2147483647 decimal.
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